Question: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 1} = \dfrac{-3x + 4}{x - 1}$
Answer: Multiply both sides by $x - 1$ $ \dfrac{x^2}{x - 1} (x - 1) = \dfrac{-3x + 4}{x - 1} (x - 1)$ $ x^2 = -3x + 4$ Subtract $-3x + 4$ from both sides: $ x^2 - (-3x + 4) = -3x + 4 - (-3x + 4)$ $ x^2 + 3x - 4 = 0$ Factor the expression: $ (x - 1)(x + 4) = 0$ Therefore $x = 1$ or $x = -4$ At $x = 1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 1$, it is an extraneous solution.